\(n_{CO_2}=\dfrac{19,8}{44}=0,45\left(mol\right)\)

\(n_{H_2O}=\dfrac{5,4}{18}=0,3\left(mol\right)\)

\(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Bảo toán C: n_C(A) = 0,45 (mol)

Bảo toàn H: _n(H(A) = 2.0,3 = 0,6 (mol)

Bảo toán O: n_O(A) + 2.n_O2 = 2.n_CO2 + n_H2O

=> n_O(A) = 2.0,45 + 0,3 - 0,6.2 = 0

n_C : n_H = 0,45 : 0,6 = 3:4

=> CTPT: (C₃H₄)_n

Mà M = 10.4 = 40(g/mol)

=> n = 1 => CTPT: C3H4

\(n_{CO_2}=\dfrac{20,16}{22,4}=0,9\left(mol\right)\); \(n_{H_2O}=\dfrac{7,2}{18}=0,4\left(mol\right)\)

Bảo toàn C: n_C(aspirin) = 0,9 (mol)

Bảo toàn H: n_H(aspirin) = 0,4.2 = 0,8(mol)

=> \(n_O=\dfrac{18-0,9.12-0,8.1}{16}=0,4\left(mol\right)\)

\(\left\{{}\begin{matrix}\%C=\dfrac{0,9.12}{18}.100\%=60\%\\\%H=\dfrac{0,8.1}{18}.100\%=4,444\%\\\%O=\dfrac{0,4.16}{18}.100\%=35,556\%\end{matrix}\right.\)

b) n_C : n_H : n_O = 0,9 : 0,8 : 0,4 = 9:8:4

=> CTHH: (C₉H₈O₄)_n

Mà M = 180g/mol

=> n = 1 => CTHH: C₉H₈O₄

c) PTHH: Ca(OH)₂ + CO₂ --> CaCO₃\(\downarrow\) + H₂O

_________________0,9---->0,9

=> m_CaCO3 = 0,9.100 = 90(g)

Xét m_CO2 - m_CaCO3 = 44.0,9 - 90 = -50,4

=> Dd sau pư có khối lượng giảm đi 50,4g

D

Gọi thể tích NaOH cần dùng là V(l)

=> nNaOH = 0,5V (mol)

n_HCl = 0,2.0,1=0,02(mol); n_H2SO4 = 0,1.0,1=0,01(mol

PTHH: NaOH + HCl --> NaCl + H₂O

_____0,02<---0,02

2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

_0,02<---0,01

=> 0,5V = 0,04

=> V = 0,08 (l)

\(n_{CaCO_3}=\dfrac{25}{100}=0,25(mol)\\ CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow\\ \Rightarrow n_{CO_2}=0,25(mol)\\ \Rightarrow V_{CO_2}=0,25.22,4=5,6(l)\)

\(n_{NaOH}=0,006\left(mol\right)\\ \Rightarrow n_{Na^+}=0,006\left(mol\right);n_{OH^-}=0,006\left(mol\right)\\ n_{H_2SO_4}=0,005\left(mol\right)\\ \Rightarrow n_{H^+}=0,01\left(mol\right);n_{SO_4^{2-}}=0,005\left(mol\right)\\ H^++OH^-\rightarrow H_2O\\ LTL:\dfrac{0,01}{1}>\dfrac{0,006}{1}\Rightarrow H^+dư\\ \left[H^+_{dư}\right]=\dfrac{0,01-0,006}{0,1}=0,04M\\ \left[Na^+\right]=\dfrac{0,006}{0,1}=0,06M\\ \left[SO_4^{2-}\right]=\dfrac{0,005}{0,1}=0,05M\)

a)

$NaOH  + HCl \to NaCl + H_2O$
$OH^- + H^+ \to H_2O$

b)

$n_{HCl} = 0,1.0,01 = 0,001(mol)$
$n_{NaOH} = 0,2.0,5= 0,1(mol)$

$\Rightarrow$ NaOH dư, HCl hết

$n_{NaOH\ pư} = 0,001 \Rightarrow n_{NaOH\ dư} = 0,1 - 0,001 = 0,099(mol)$

$\Rightarrow [OH^-] = \dfrac{0,099}{0,1 + 0,2} = 0,33M$
$\Rightarrow pOH = -log(0,33) = 0,48 $

$pH = 14 - pOH = 14 - 0,48 = 13,52$

\(2KOH+H_2SO_4->K_2SO_4+2H_2O\)

\(OH^-+H^+->H_2O\)

\(n_{H_2SO_4}=0,05.0,2=0,01\left(mol\right);n_{KOH}=0,2.0,1=0,02\left(mol\right)\)

PTHH: \(2KOH+H_2SO_4->K_2SO_4+2H_2O\)

_____0,02------->0,01

=> KOH, H2SO4 phản ứng vừa đủ, tạo ra dd K₂SO₄

=> pH = 7